As to the other question, there is no uniform constant for all $0<\alpha<1$, and the reason is that the above constant is indeed optimal. Precisely, for any iteration $\gamma_{k+1}= \gamma_k(1- c \gamma_k^\alpha)$ with $1-c\gamma_0^\alpha >0$ we have
$$\gamma_k=Ck^{-1/\alpha}\big(1+o(1)\big)\, ,\quad\mathrm{for}\, k\to\infty \, .$$

I don't see how you estimate $\gamma_0$ in terms of $c,\alpha$ in the endgame, but otherwise I like it.
–
GH from MOAug 2 '11 at 14:47

It's just the initial assumption in the OP, $1−c\gamma_k^\alpha > 0$ for all $k$, which indeed is ensured by $1−c\gamma_0^\alpha > 0$, and it is necessary for $\gamma_k$ to be well-defined for all $k$ (I've added the remark).
–
Pietro MajerAug 2 '11 at 16:15

I always obtain decay rates for iterative inequalities like that by comparing them with differential equations.

The particular difference equation you give can be compared to $\dot y=-cy^{1+\alpha}$.
If you think of the right side as a "speed" then the speed is $cy^{1+\alpha}$ moving to the left and decreases over time. On the other hand if you think of the difference equation as moving to the left at speed $y^{1+\alpha}$ which stays constant for 1 unit of time, it is clear that the difference equation approaches 0 faster.

In other words, the solution to the differential equation is an upper bound for the solution to the difference equation. If you work a bit harder you can find another differential equation that serves as a lower bound for the difference equation so that you can get good bounds from above and below.